I need some help with this question:
Consider an individual who possesses the Bernoulli utility function of $u(x)=\dfrac{x^{1-\gamma }}{1-\gamma }$ where $\gamma>0$, $\gamma \neq 1$. Who maintains an initial level of wealth $W$. Suppose that she is considering a lottery where she wins or losses $x$ with equal probability. Let $p$ be the amount the agent is just willing to pay to avoid the lottery. Find this value, and show that it is decreasing in wealth.
My work:
Let $W-p$ denote certainty equivalent.
So, $u(W-p)= \dfrac{1}{2}u(W-x)+\dfrac{1}{2}u(W+x)$
$\Rightarrow \dfrac{(W-p)^{1-\gamma }}{1-\gamma}= \dfrac{1}{2}\dfrac{(W-x)^{1-\gamma}}{1-\gamma}+\dfrac{1}{2}\dfrac{(W+x)^{1-\gamma}}{1-\gamma}$
After doing some algebraic manipulations, I am left with this equation:
$log(W-p)=\dfrac{-2log(2)+log(W-x)+log(W+x)}{1-\gamma}$
From here how do I find $p$?
If you have an equation of the form
$$ \log(W - p) = \frac{a\log b + c\log d + e\log f}{g}$$
then you should solve this equation for $p$ by getting rid of the $\log$s. The easiest way to do this is boil the RHS down to one $\log$ by combining terms using the rules $p\log q$ = $\log (q^p)$ and $\log p + \log q$ = $log (pq)$. Hence
$$ \log(W - p) = \frac1{g} \left(\log (b^a d^c f^e)\right) = \log \left(\left( b^a d^c f^e \right)^{1/g}\right) $$
You can solve this by taking exponents of both sides, which eliminates the $\log$s.
However, you don't need to do this. Starting with your equation
$$\frac{(W-p)^{1-\gamma }}{1-\gamma}= \frac{1}{2}\frac{(W-x)^{1-\gamma}}{1-\gamma}+\frac{1}{2}\frac{(W+x)^{1-\gamma}}{1-\gamma}$$
first cancel through the common fractions
$$(W-p)^{1-\gamma } = \frac{1}{2}\left((W-x)^{1-\gamma}+(W+x)^{1-\gamma}\right)$$
then take raise each side to the power $\dfrac1{1-\gamma}$, leaving $W-p$ on the left (the expression on the right doesn't simplify unfortunately). You can then do a simple rearrangement to find $p$.
PS your "after some more algebraic calculations" looks wrong to me. I don't see how you get to your final equation. Did you try to put $\log(a+b)=\log a + \log b$ somewhere?